1. ## Vector differential identities

Could someone help me prove this identity:

Let F and G be vector fields and let % be the operator (d/dx)*i + (d/dy)*j +(d/dz)*k where d/dx equals the partial derivative of x, etc.

%*(FxG) = (%xF)*G - F*(%xG).
* is the dot product
x is the cross product
- is subtraction

I'm sorry for the notation. Thanks.

2. Originally Posted by USCGuy
Could someone help me prove this identity:

Let F and G be vector fields and let % be the operator (d/dx)*i + (d/dy)*j +(d/dz)*k where d/dx equals the partial derivative of x, etc.

%*(FxG) = (%xF)*G - F*(%xG).
* is the dot product
x is the cross product
- is subtraction
I assume you have,
$\bold{F},\bold{G}$
Are vector fields.
And you want to show,
$\nabla \cdot (\bold{F}\times \bold{G} )=\mbox{div } (\bold{F}\times \bold{G})$
And,
$(\nabla\times \bold{F})\cdot \bold{G}-(\nabla \times \bold{G})\cdot \bold{F}=(\mbox{curl }\bold{F})\cdot \bold{G}-(\mbox{curl }\bold{G})\cdot \bold{F}$
Thus, you need to show,
$\mbox{div } (\bold{F}\times \bold{G})=(\mbox{curl }\bold{F})\cdot \bold{G}-(\mbox{curl }\bold{G})\cdot \bold{F}$
I do not want to type of this stuff out.
But consider you have in general,
$\bold{F}(x,y,z)=u_1(x,y,z)\bold{i}+u_2(x,y,z)\bold {k}+u_3(x,y,z)\bold{k}$
$\bold{G}(x,y,z)=v_1(x,y,z)\bold{i}+v_2(x,y,z)\bold {k}+v_3(x,y,z)\bold{k}$.
(And they are well-behaved).

Now, just substitute that into both sides of the equation and show that they are true. I just do not want to do that because it is time consuming to type. It is really and easy proof, just substitute and see that the right hand and left hand match.

3. It just appears that after doing everything out, I get double what I should get.

Instead of d/dx(F2G3-F3G2) I keep getting d/dx(2(F2G3)-2(F3G2)) and the same for d/dy and d/dz. It makes sense except for this.

Does something equal zero?